I already have the year being 2.2 earth years(800 earth days to be exact) on Kepler Bb due to transitional phases(of which there are 4 and each one lasts 80 earth days) and the main seasons being 1/3 longer than ours.

Now I have decided that the day should be 8 hours longer than ours. This leads to the year being another 200 days longer(so 1000 earth days per Kepler Bb year).

Likewise I could have each hour be 20 minutes longer. This would add yet another 250 earth days according to my calculations.

As you can probably see, this going backwards to smaller and smaller units and then back forward is not a good way to increase time.

Here are my calculations: Day calculations:

800 earth days per Kepler Bb year * 8 extra hours per Kepler day = 6400 extra earth hours

6400 extra earth hours

--------------------------------------------- = 200 extra Kepler days assuming hour is the same

32 earth hours per Kepler day

Hour calculations:

32 earth hours per Kepler day * 20 extra minutes per Kepler hour = 640 extra minutes per Kepler day

640 extra minutes per Kepler day * 1000 days per Kepler Bb year = 640,000 extra minutes per Kepler Bb year

640,000 extra earth minutes per Kepler Bb year

------------------------------------------------------------------- = 8,000 extra Kepler hours

80 minutes per Kepler hour

8,000 extra Kepler hours

------------------------------------------- =250 extra Kepler days assuming minute is the same

32 Kepler hours per Kepler day

So that is a total of 1250 Kepler days. In earth years that is 3.4 earth years. Yet another 1.2 earth years longer than the 2.2 earth years assuming day is the same.

Now I could go through this whole thing again with seconds per Kepler minute and then years per Kepler x(arbitrary time unit longer than year) and Kepler x per Kepler y etc. but I'd rather not(lots and lots of algebra using that method).

However, I want to make a webpage(I am using notepad++ for HTML/CSS/JS + Other). With the notepad++ I can essentially create my own local network website for free. Anyway, this website I plan to be all about Kepler Bb and the life and civilization on that planet.

In order to do that I might as well make every time unit except seconds different. Sorry if the fractions don't look good.

Anyway, I want a good method for doing these calculations. Sure I can keep converting between earth and Kepler and between Kepler and Kepler but if I don't have to, then I don't want to.

So do you have any idea of what a good method would be?

  • $\begingroup$ A year is a single orbit around the sun, while a day is a single rotation of the planet. So why would adding 8 hours to the day add time to the year? (i.e. change the time of the orbit). It would be slower rotation so just less days in the year. But if you really want to extend the year, have you taken into account the planet needs to be further away from the star and so the star has to be bigger/hotter to sustain life? $\endgroup$
    – komodosp
    Commented Jul 20, 2016 at 10:20
  • $\begingroup$ The way I see it, I am slowing the rotation rate but keeping the revolution rate the same. If I keep the revolution rate the same, then adding time to smaller time units adds time to the year both in total time(seconds) and in units $\endgroup$
    – Caters
    Commented Jul 20, 2016 at 14:27
  • $\begingroup$ Or rather slowing the revolution rate accordingly so that time is extended in all units(earth and Kepler). Also yes I did take that into account because all planets in the solar system Kepler Bb is in orbit around a 4 star star system(close double binary to be precise). Each individual star is outside the type 2 supernova mass range and because they are close, I can consider them as just 1 big star when it comes to things like habitable zone calculations. $\endgroup$
    – Caters
    Commented Jul 20, 2016 at 14:37

3 Answers 3


If you're doing these calculations by hand, then my favorite "go-to" answer is to input it all into a spreadsheet in a manner that works for you (though I'd very much advise keeping the column & row headers named, and include copious comment-notes for anything that you do).

If you're already doing something of the sort, then I'd have to go with the suggestion that I often hear from a programming friend of mine: "Why not just code it?". For me, this isn't the best answer, since I'm comfortable with spreadsheets and VBA (as needed), but you might find C++, Java, etc. to be of more use.


I left a comment originally, but I decided to turn it into an answer because really, that's an answer. I did do that sort of thing in Excel, JavaScript, C# and Java. The super simple way I do it follows.

Consider these variables: SS the number of seconds. MM the number of minutes. HH the number of hours. sm the number of seconds in a minute. mh the number of minutes in an hour.

First step is to convert your input into seconds. Could be a string, array or object, doesn't matter, as long as you can get your SS, MM and HH values. The result of that will be:

$$ x = \text{SS} + \text{MM} \times \text{sm} + \text{HH} \times \text{mh} \times \text{sm} $$

Then you convert it into the relevant time system.

$$ \text{SS} = x \mod \text{sm} $$

$$ \text{MM} = \left( \frac{x - \text{SS}}{\text{sm}} \right) \mod \text{mh} $$

$$ \text{HH} = \frac{x - \text{SS} - \text{MM} \times \text{sm}}{\text{sm} \times \text{mh}} $$

Those calculation should work. In the case they don't, woops.

And then you're done. You just transform Earth time into seconds, then transform second into not-Earth time. Factoring in days and years will work in a similar way, if you really can't figure it I could add them later.

Little thing to consider: there are stuff dependent on time. Speed for instance is one. If you change the duration of a second it changes the value for speed of light in vacuum. 50kph won't be the same on Earth than on Kepler, be sure to remember that when distributing speeding tickets.

  • $\begingroup$ I will be sure to remember that speed is dependent on time. And I figured I might as well keep the second constant throughout the whole universe regardless of other differences between time units. That would be very important if at some point there is interplanetary communication not just within a solar system but within a whole galaxy. It would also be important from the physics perspective(though I don't really delve into physics all that much). $\endgroup$
    – Caters
    Commented Jul 20, 2016 at 18:08

Time is an illusion. Lunchtime doubly so.

The values of 24 and 60 are very human values. It's important to remember this when you're considering how you manipulate your clock.

24 will easily divide by 2,3,4,6,8,12.
60 will easily divide by 2,3,4,5,6,10 etc.

If you're used to doing calculations in your head then you find such numbers much easier to work with than a decimal system as you're often only dealing with integer values and not fractions.

The concept of a day is also a very human thing. You get up in the morning when it's light and go to bed in the evening when it gets dark. The day is of a duration that's easily divided up. It's important to note that before time zones and clocks, that the day was 12 hours and the night was 12 hours and that was an end to the matter. The fact that those hours varied in length through the year was a minor detail.

Which brings me to my point: It doesn't matter. A day is a day, that's what most people think in, days and nights. You don't need to keep converting between Earth time and Kepler time unless you have instantaneous communications between the two. Once you have defined your base relative times you never need to run the calculation again.

It's only when there's cause on on planet and effect on the other that it actually matters what the relative times are. The lightcone says that for most day to day purposes they might as well be in different universes.


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